A Ferris wheel is a vertical, circular amusement ride with radius 9 m. Riders si

Question

A Ferris wheel is a vertical, circular amusement ride with radius 9 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 8.5 s. Consider a rider whose mass is 50 kg.

At the bottom of the ride, what is the rate of change of the rider's momentum?

At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider?

At the bottom of the ride, what is the vector force exerted by the seat on the rider?

Next consider the situation at the top of the ride. At the top of the ride, what is the rate of change of the rider's momentum?

At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider?

At the top of the ride, what is the vector force exerted by the seat on the rider?

Explanation / Answer

given that

r =9 m

t = 8.5 s

m = 50 kg

w = 2*pi / t = 2*3.14 / 8.5 = 0.738 rad/s

(a)

At the bottom of the ride, rate of change of the rider's momentum,

dp / dt = Fc = m*w^2*r

dp / dt = 50*(0.738)^2*9 = 245.63 kg*m/s^2

(b)

vector gravitational force exerted by the Earth on the rider,

F = -m*g = -50*9.81

F = -490.5 N

(c)

vector gravitational force exerted by the Earth on the rider is,

N = m*w^2*r + m*g

N = 245.63 + 490.5 = 736.13 N

(d)

At the top of the ride, rate of change of the rider's momentum,

dp / dt = Fc = 245.63 kh*m/s^2

(e)

Fg = m*g = 490.5 N

(f)

At the top of the ride, vector force exerted by seat on the rider,

N = m*w^2*r - m*g = 245.63 - 450.9

N = -205.27 N

answer

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